Answer:
B - Only a concentration of mass in the nucleus could explain the deflection of the incident particles.
A car travels first 10 km in 20 minutes and another 10 km in 30 minutes. What is the average speed of the car in m/s?
Total distance = 10 km + 10 km = 20 km
1 km = 1000 m
20km x 1000 = 20,000 m
Total time = 20 min. + 30 min. = 50 minutes
Average speed = Distance / time
Average speed = 20,000/50 min
Average speed = 400 m/s
A block of amber is placed in water and a laser beam travels from the water through the amber. The angle of incidence is 35 degrees while the angle of refraction is 24 degrees. What is the index of refraction of amber?
(index of refraction of water is 1.33)
maize is a monocotyledonous seed and pea is a dicotyledonous seed why? give short and the suitable answer I will mark you as a brainelist
Answer:
A dicot is a flowering plant that has one seed leaves. The monocot plants have a single cotyledon. Maize only has one cotyledon in their seed, so it's a monocot. Seeds having two Cotyles are mainly called a Dicot. A pea is a dicotyledonous plant, the seed (the pea itself) has two halves, cotyledons, hence dicot being 2.
Explanation:
One or more of the cotyledons are the first to appear from a germinating seed. Based on the number of cotyledons, botanists classify flowering plants (angiosperms) into :
a) plants with one embryonic leaf, termed monocotyledonous (monocots).
b) plants with two embryonic leaves, termed dicotyledonous (dicots).
Helpful Link:
https://www.vedantu.com/question-answer/in-pea-caster-and-maize-the-number-of-cotyledons-class-11-biology-cbse-5f626a17e5bde9062ff6d2a3
A 5.41 kg ball is attached to the top of a vertical pole with a 2.37 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.75 m/s in a horizontal circle with the string remaining taut. Calculate the angle θ, between 0∘ and 90∘, that the string makes with the pole. Use g=9.81 m/s2.
Answer:
θ = 66º
Explanation:
This exercise of Newton's second law must be solved in part, let's start by finding the slowing down acceleration of the ball
a = v² / r
the radius of the circle is
sin θ = r / L
r = L sin θ
we substitute
a = v² /L sin θ
now let's write Newton's second law
vertical axis
T_y -W = 0
T_y = W
radial axis
Tₓ = m a (1)
let's use trigonometry for the components of the string tension
cos θ = T_y / T
sin θ = Tₓ / T
Tₓ = T sin θ
we substitute in 1
T sin θ = [tex]\frac{m \ v^2}{L \ sin \theta}[/tex]
T L sin² θ = m v²
we write our system of equations
T cos θ = m g
T L sin ² tea = m v²
we divide the two equations
L [tex]\frac{sin^2 \theta}{cos \theta}[/tex] = v² / g
(1 -cos²)/ cos θ = [tex]\frac{v^2 }{g \ L}[/tex]
1 - cos² θ = [tex]\frac{4.75^2}{9.81 \ 2.37}[/tex] cos θ
cos² θ + 0.97044 cos θ -1 = 0
we change variable cos θ = x
x² + 0.97044 x - 1 =0
x= [tex]\frac{-0.97 \pm \sqrt{0.97^2 - 4 1} }{2}[/tex]
since the square root is imaginary there is no real solution to the problem, suppose that the radius is 1 m r = 1 m
T sin θ = [tex]\frac{m \ v^2}{ r}[/tex]
T cos θ = m g
resolved
tan θ = [tex]\frac{v^2}{ r g}[/tex]
θ = tan⁻¹ ( 4.75²/ 1 9.81)
θ = 66º
There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12 V until it is loaded completely. Then the capacitor is disconnected from the voltage source. After this the two plates of the capacitor are driven apart until their distance is 5 mm. Now a positive test charge of 1 nC is brought from the positively charged plate to the negatively charges plate. How large is the kinetic energy of the test charge? The test charge of 1 nC can be regarded to be so small that it does not influence the electric field between the two plates of the capacitor.
The kinematic energy of the positive charge is 2 10⁻⁸ J
This electrostatics exercise must be done in parts, the first part: let's start by finding the charge of the capacitor, the capacitance is defined by
C = [tex]\frac{Q}{\Delta V}[/tex]
C = ε₀ [tex]\frac{A}{d}[/tex]
we solve for the charge (Q)
[tex]\frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}[/tex]
indicates that for the initial point d₁ = 3 mm = 0.003 m and the voltage is DV₁ = 12
Q = [tex]\epsilon_o \ \frac{A \ \Delta V_1 }{d_1}[/tex]
Now the voltage source is disconnected so the charge remains constant across the ideal capacitor.
For the second part, the condenser is separated at d₂ = 5mm = 0.005 m
Q = \epsilon_o \ \frac{A \ \Delta V_2 }{d_2}
we match the expressions of the charge and look for the voltage
[tex]\frac{\Delta V_1}{d_1} = \frac{\Delta V_2}{d_2}[/tex]
ΔV₂ = [tex]\frac{d_2}{d_1 } \ \Delta V_1[/tex]
The third part we use the concepts of conservation of energy
starting point. With the test load (q = 1 nC = 1 10⁻⁹ C) next to the left plate
Em₀ = U = q DV₂
Em₀ = q \frac{d_2}{d_1 } \ \Delta V_1
final point. Proof load on the right plate
Em_f = K
energy is conserved
Em₀ = em_f
q \frac{d_2}{d_1 } \ \Delta V_1 = K
we calculate
K = 1 10⁻⁹ 12 [tex]\frac{0.005}{0.003}[/tex]
K = 20 10⁻⁹ J
In this exercise, as the conditions at two different points of separation give, the area of the condenser is not necessary and with conservation of energy we find the final kinetic energy of 2 10⁻⁸ J
can someone pls help me calculate this?
Answer:
10N is the answerrrerreer
Answer:
f= 100N
Explanation: F=m×(v₀-vf/t)
=0.05ₓ(200-0/0.1)
=0.05ₓ2000
=100N
please mark as brainliest
A long distance runner running a 5km track is pacing himself by running 4.5km/h at 9km/h and the rest at 12.5km/h9
Complete Question:
A long distance runner running a 5.0km track is pacing himself by running 4.5km at 9.0km/hr and the rest at 12.5km/hr. What is the average speed?
Answer:
Average speed = 9.7333 km/h
Explanation:
Let the total distance be divided into A and B.Given the following data;
Total distance = 5 kmDistance A = 4.5 kmSpeed A = 9.5 km/hrSpeed B = 12.5 km/hrTo find the average speed;
First of all, we would determine the time taken to cover distance A in speed A by using the formula;
[tex] Time \ A = \frac {Distance \; A}{Speed \; A} [/tex]
Substituting the values into the formula, we have;
[tex] Time \ A = \frac {4.5}{9.5} [/tex]
Time A = 0.4737 hours
Total distance = distance A + distance B
5 = 4.5 + distance B
Distance B = 5 - 4.5
Distance B = 0.5 Km
Next, we would determine the time to cover distance B in speed B;
[tex] Time \ B = \frac {0.5}{12.5} [/tex]
Time A = 0.04 hours
Total time = time A + time B
Total time = 0.4737 + 0.04
Total time = 0.5137 hours
Now, we would solve for the average speed;
Mathematically, the average speed of an object is given by the formula;
[tex] Average \; speed = \frac {total \; distance}{total \; time} [/tex]
[tex] Average \; speed = \frac {5}{0.5137} [/tex]
Average speed = 9.7333 km/h
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. You reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 10 \mathrm { m } / \mathrm { s } ^ { 2 }10m/s 2 . a. How much distance is between you and the deer when you come to a stop
Answer:
Explanation:
Discount the time here; it's not important. It doesn't tell you how long it takes the car to stop, it only refers to reaction time, which means nothing in the scheme of things.
The useful info is as follows:
initial velocity = 20 m/s
final velocity = 0 m/s
a = -10 m/s/s
and we are looking for the displacement. Use the following equation:
[tex]v^2=v_0^2+2a[/tex]Δx
where v is the final velocity, v₀ is the initial velocity, a is the deceleration (since it's negative), and Δx is displacement. Filling in:
[tex]0^2=(20)^2+2(-10)[/tex]Δx and
0 = 400 - 20Δx and
-400 = -20Δx so
Δ = 20 meters
The pressure at the ice point for a constant-volume gas thermometer is 4.81 x 10^4Pa.
While that at the steam point is 6.48 x 10^4 Pa.
What pressure would the thermometer indicate at 50°c?
Answer:
0 deg C = 4.81E4 pressure at 0 deg
100 deg C = 6.48E4 pressure at steam point
100 deg C - 50 deg C = (6.48 - 4.81) * 10^4 = 1.67E4 Pa
50 deg C = 50 / 100 * 1.67E4 + 4.81E4 = 5.65E4 Pa Just the halfway point between the two given pressures
What is an effect of continental drift?
Answer: An effect of continental drift is causing tectonic plates resting upon the convecting mantle to move which results in natural disasters like earthquakes, volcanic eruptions, and more.
3kg of water at 80degree celcius is added to 8 kg of water at 25 degree celcius. find the temperature of final mixture provided there is no loss of heat in the surrounding. the specific heat capacity is 4200j/kg
Answer:
hope fully it help s
describe how air resistance would affect a falling object
Answer:
With air resistance, acceleration throughout a fall gets less than gravity (g) because air resistance affects the movement of the falling object by slowing it down. How much it slows the object down depends on the surface area of the object and its speed
Explanation:
PLZ MARK AS THE BRAINLIEST
HAVE A GOOD DAY
: )
The refractive index of glass is 1.52
and that of air is 1.00. Draw a labelled diagram to show how a light ray bends when it travels from glass to air.
Explanation:
..upper side is glass ..
GUYS PLEASE HELP ME THIS IS FOR MY FINAL !!!
Answer:
just add co-ratios and then find the average.
Explanation:
a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone
Answer:
The centripetal acceleration of the stone is 5 m/s²
Explanation:
The length of the string to which the stone is attached, r = 1 m
The speed with which the string is rotated, v = 5 m/s
The centripetal acceleration, [tex]a_c[/tex], is given as follows;
[tex]a_c = \dfrac{v^2}{r}[/tex]
Therefore, the centripetal acceleration of the stone found as follows;
[tex]a_c = \dfrac{(5 \ m/s)^2}{1 \ m} = 5 \ m/s^2[/tex]
The centripetal acceleration of the stone, [tex]a_c[/tex] = 5 m/s².
State one effect of increase in heat for a temperature range of 50°C to 100°C
on Plastic
Answer:
why is fraction called a necessary evil
The heat for a temperature in plastic increases at the range of 50°C to 100°C then the plastic decreases its stiffness or flexural modulus.
It is given that the temperature is increased from 50°C to 100°C that means temperature is increasing.
It is required to state the effect of increase in heat on plastic.
What will be the effect of increase in heat for a temperature range of 50°C to 100°C on Plastic?The plastic starts to lose its stiffness or we say that it converts into a softened material as long as we increase the temperature of the material.
So if we increase the temperature too much or exceed above the limit of temperature range then it will distort.
Also there are many qualities that can be affected by increasing the temperature like mechanical property, material fatigue or chemical phenomenon.
Therefore, the heat for a temperature range of 50°C to 100°C
Plastic increases then the plastic decreases its stiffness or flexural modulus.
Learn more about the range of temperature here:
https://brainly.com/question/11316625
#SPJ5
find the expression for pressure exerted by fluid with proper description
Answer: Let us calculate the pressure exerted on the bottom by the weight of the fluid. That pressure is the weight of the fluid mg divided by the area A supporting it (the area of the bottom of the container): P=mgA P = m g A .
Explanation:
Does water exist on Mars? Explain your answer. Why didn't Jupiter become a star during the early.
Answer:
yes water was discovered in mars
The atomic bomb dropped on Hiroshima converted about 7.00x10-4kg of mass to energy. How much energy did that bomb produce?
A)2.10x10^5J
B)7.78x10^-21J
C)6.30x10^13J
D)2.10x10^61J
Answer:
[tex] \sf \: given \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bf \: mass \: \: \: m \: = 7.00 \times {10}^{ - 4} \: kg \\ \\ \bf \: E=mc^2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = > E=7.00 \times {10}^{ - 4} \times ({3 \times {10}^{8} })^{2} \\ \\ = > \green{ \boxed{ E = 6.3 \times {10}^{13} \: J}}[/tex]
a motorcycle covers 500 meters in 25 seconds. calculate the average velocity
Vtb xe máy là:
v=s : t= 500 : 25=20(m/s)
determjne the density of liquid whose relative density is 1.25 given that the density is 1000kgm-3
Answer:
divide the density of solution by density of water
EXPLANATION:
LIKE:
1.25÷1000kgm-3
which physical property of the gas molecules gives the measurement of temperature
The temperature of a gas molecule is measured by the average translational kinetic energy
Answer:
The temperature of a gas is a measure of the average translational kinetic energy of the molecules. In a hot gas, the molecules move faster than in a cold gas; the mass remains the same, but the kinetic energy, and hence the temperature, is greater because of the increased velocity of the molecules
Explanation:
This is also from Go0gle because my explanation would've been an essay long .
but in shorter version if the gas molecules move fast it's hot an if it moves slow its cold hope this helps .
Hideki had normal vision for most of his life, but now that he is in his 60s, he has started to have difficulty focusing on near objects. He went to an optometrist, who explained that his vision problem was the result of the lenses in his eyes losing elasticity due to aging. Which condition does Hideki have
Answer:
He is suffering from hypermetropia.
Explanation:
There are some defects of vision.
Longsightedness of hypermetropia : It is the defect of vision in which the person is not able to see the nearby objects clearly but can see the far off objects clearly. This is due to the elongation of size of eye ball. It is cured by using convex lens of suitable focal length.
Nearsightedness or Myopia : It is the defect of vision in which a person is not able to see the far off objects clearly but can see the nearby objects clearly. It is due to the contraction in the size of eye ball. It is cured by using concave lens of suitable focal length.
So, Hideki is suffering from hypermetropia. So, he should use the convex lens of suitable focal length.
Diffraction spreading for a flashlight is insignificant compared with other limitations in its optics, such as spherical aberrations in its mirror. To show this, calculate the minimum angular spreading in rad of a flashlight beam that is originally 5.90 cm in diameter with an average wavelength of 610 nm.
Answer:
The answer is "[tex]1.2566 \times 10^{-5}\,rad[/tex]".
Explanation:
As per the Rayleigh Criterion the minimum angular spreading, for a circular aperture:
[tex]\theta_{\mathrm{min}}\approx \sin\theta=1.22\,\frac{\lambda}{d}[/tex]
[tex]\theta_{\mathrm{min}}=\mathrm{1.22\,\frac{\left( 610\,nm \right)}{\left( 5.90\,cm \right)}=1.22\,\frac{\left( 610\times10^{-9}\,m \right)}{\left( 5.90\times10^{-2}\,m \right)}}[/tex]
[tex]=1.22\times 103.389 \times 10^{-7}\\\\=1.22\times 1.03 \times 10^{-5}\\\\=\mathrm{1.2566 \times 10^{-5}\,rad}[/tex]
A conversion factor is a ratio of ____ measures
Opposite
Larger
Smaller
Equivalent
Answer:
Equivalent
A conversion factor is a ratio of equivalent measures
b) A force is represented in magnitude and direction as (6N, 250degrees. Find both the vertical and horizontal components of the force.
Answer:
Explanation:
To find the horizontal component, the x component specifically, use the formula:
[tex]V_x=Fcos\theta[/tex] and for the vertical component, the y component, use the formula:
[tex]V_y=Fsin\theta[/tex]
where F is the magnitude of the force and theta is the angle in degrees.
For the x-component:
[tex]V_x=6cos250[/tex] so
[tex]V_x=-2.1[/tex] and depending upon whether this is a displacement vector or a velocity vector, the label would be meters/feet or m/s, respectively.
For the y-component:
[tex]V_y=6sin250[/tex] so
[tex]V_y=-5.6[/tex]
A 117 kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 62.5 kg person stands on the platform at a distance of 1.05 m from the center, and a 28.3 kg dog sits on the platform near the person 1.43 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.
Answer:
I_syst = 278.41477 kg.m²
Explanation:
Mass of platform; m1 = 117 kg
Radius; r = 1.61 m
Moment of inertia here is;
I1 = m1•r²/2
I1 = 117 × 1.61²/2
I1 = 151.63785 kg.m²
Mass of person; m2 = 62.5 kg
Distance of person from centre; r = 1.05 m
Moment of inertia here is;
I2 = m2•r²
I2 = 62.5 × 1.05²
I2 = 68.90625 kg.m²
Mass of dog; m3 = 28.3 kg
Distance of Dog from centre; r = 1.43 m
I3 = 28.3 × 1.43²
I3 = 57.87067 kg.m²
Thus,moment of inertia of the system;
I_syst = I1 + I2 + I3
I_syst = 151.63785 + 68.90625 + 57.87067
I_syst = 278.41477 kg.m²
2. Speed limits on curves help to reduce the effect of ........force on a vehicle
A. enertia
b. centrifugal
c. gravity
Answer:
enertia
Explanation:
enertia
The correct answer is (B) Centrifugal Force
The centrifugal force on a vehicle on the curves is outwards or can say away from the curve and pulls the out of the track.
the centrifugal force is given by [tex]F=\frac{mv^{2} }{r}[/tex], which is directly proportional to speed v.
Speed limits on the curves help in reducing the centrifugal force by reducing the speed.
Learn more about centrifugal force:
https://brainly.com/question/25546106
How does 'g' vary from place to place?
Explanation:
The acceleration g varies by about 1/2 of 1 percent with position on Earth's surface, from about 9.78 metres per second per second at the Equator to approximately 9.83 metres per second per second at the poles.
explain how force and surface area affect the applied pressure.
Answer:
force and surface area are two factors affecting pressure on solids
more the force you apply, more will be the pressure
pressure and force are directly proportional meaning if Force is greater, pressure will also be greater
more the surface area of the solid less will be the pressure
surface area and pressure are inversely proportional meaning if surface are is big, pressure will be less, surface area small, pressure will be greater
Answer and Explanation:
We have a basic equation: Pressure = Force/Area.
So for example:
Increase pressure - increase the force or reduce the area the force acts on.
Decrease pressure - decrease the force, or increase the area the force acts on
The force per unit area is pressure. The force on the object is spread over the surface area. The area where the force is applied is divided by the equation for pressure.
Cheers,
